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I have a problem in understanding a proof of Goldstine–Weston density theorem. The only thing I don't know in the proof is the part of Helly's theorem to be related. The Goldstine–Weston density theorem is as follows:

Suppose that $X$ is a Banach space and $J_X:X\to X^{\star\star}$ is the canonical embedding. Then $\overline{J_X(X)}^{w^{\star}} = X^{\star\star}$.

Proof:

We need only to prove $~\overline{J_X(U(X))}^{w^{\star}} = U(X^{\star\star})$, where $U(X)=\{ x\in X : \| x \|\leq 1 \}$, $U(X^{\star\star})=\{ x^{\star\star} \in X^{\star\star}: \| x^{\star\star} \|\leq 1 \}$. Let $\Phi \in U(X^{\star\star})$, for any $w^{\star}$ neighbourhood $V$ of $\Phi$ defined by \begin{equation} V=\{ \Psi \in X^{\star\star}: |\left<x^{\star}_{i}, \Psi - \Phi\right>| < \varepsilon, x^{\star}_{i}\in X^{\star}, i=1, \cdots, n \}, \end{equation} and let $r=\max\{ \| x^{\star}_{i} \|: i=1, \cdots, n \}$. $\color{magenta}{\text{By Helly's theorem,}}$ there exists $x_{\varepsilon} \in X$ such that $\| x_{\varepsilon} \| \leq 1+\frac{\varepsilon}{2r}$, $\left<x_{\varepsilon}, x^{\star}_{i}\right> = \left<x^{\star}_{i}, \Phi \right>, i=1, \cdots, n$. Set $x_{0} = \frac{r}{r + \varepsilon/2}x_{\varepsilon}$, then we have $\| x_0 \|\leq 1$ and \begin{equation} |\left< x^{\star}_i, \widehat{x}_0 - \Phi \right>| = |\left<x_0-x_{\varepsilon}, x^{\star}_{i} \right>| \leq \frac{\varepsilon}{2} \| x_0 \| < \varepsilon, \end{equation} where $\widehat{x}_0 \overset{def}{=} J_X(x_0) \in V$. Thus, $\Phi \in \overline{J_X(U(X))}^{w^{\star}}$. That is, $U(X^{\star\star}) \subseteq \overline{J_X(U(X))}^{w^{\star}}$. The reversed inclusion relation is due to Alaoglu's theorem.

My question:

I don't know what's the Helly's theorem mentioned in the proof and also I can't find this theorem in my book anywhere. Thanks in advance.

Edit $1$:

Tomek Kania has given the Helly's theorem which I want to know at first (I still want to know that from which book or place a proof of it can be found ). I have verified that there exists an element $x_{\varepsilon} \in X$. However, I can't prove the inequality $\| x_{\varepsilon} \| \leq 1+\frac{\varepsilon}{2r}$.

Tomasz Kania
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Travis Wang
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1 Answers1

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This is what is often called Helly's theorem.

Let $X$ be a Banach space. Let also $f_1, \ldots, f_n\in X^*$ and scalars $\alpha_1, \ldots \alpha_n$ be given. Then the following conditions are equivalent:

  • there exists $x\in X$ such that $\langle x, f_i \rangle = \alpha_i$ ($i\leqslant n$)
  • there exists $\gamma \geqslant 0$ such that for any scalars $\beta_1, \ldots, \beta_n$ we have $$|\sum_{i=1}^n \beta_i \alpha_i| \leqslant \gamma\cdot\|\sum_{i=1}^n \beta_i f_i \|.$$

I think that the fact you need is the following weaker version of the above theorem:

Let $X$ be a Banach space and let $x^{**}\in X^{**}$ be a unit vector. Given a finite set $F\subset X^*$ and $\varepsilon>0$, there exists $x\in X$ such that $\|x\|\leqslant 1+\varepsilon$ and $\langle x,f\rangle = \langle f, x^{**}\rangle$ for all $f\in F$.

Proof. Let $Z = \bigcap_{f\in F}\ker f$. Note that $X/Z$ is finite-dimensional (hence reflexive). Denote by $q\colon X\to X/Z$ the canonical quotient map. Choose now $x\in X$ sitting in $q^{**}(x^{**}) + (X/Z)^{**} = q^{**}(x^{**}) + X/Z$ that has norm not exceeding $1+\varepsilon$. $\square$

Note that the Principle of Local Reflexivity is a powerful extension of this theorem.

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Tomasz Kania
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  • I don't know why $\langle x,f\rangle = \langle f, x^{\star\star}\rangle$ for the $x^{\star\star}$ in it may not just right be the double dual of $x$. In the above comments, I first want to prove that $X/Z$ is finite-dimensional. It needs the property of linearity of $g$ but I can't verify this. By your comment, I have known this inequality $|x|\leq 1+\varepsilon$ now. – Travis Wang Oct 18 '14 at 13:13
  • Kernels of bounded linear functionals have codimension one so the intersection of kernels of $n$ such functionals has codimension at most $n$ (it is equal to $n$ if and only if they are linearly independent). What is a "double of dual of $x$"? $x^{}$ is just any norm-one element of $X^{}$. – Tomasz Kania Oct 18 '14 at 13:25
  • I mean $\langle x,f\rangle = f(x)=\langle x^{\star\star}, f \rangle$. But this "$x^{\star\star}$" may not be the $x^{\star\star}$ we first choose in a ball of $X^{\star\star}$. – Travis Wang Oct 18 '14 at 13:55
  • Sorry, I must be missing something. In your notation, consider $x^{**} = \Phi$. – Tomasz Kania Oct 18 '14 at 14:17
  • Sorry, I am not that mean. Only $\langle x,f\rangle = \langle f, \Phi\rangle$ is still a question in my mind. Let me say what I have thought: Since $q(\Phi)$ is an element in $X/Z$, for any $\varepsilon>0$ there exists $x\in X$ such that $|x|<|q(\Phi)|+\varepsilon$. Since $|q**(\Phi)|\leq |q|\leq 1$, we have $|x|< 1+\varepsilon$. But I can't see why $\langle x,f\rangle = \langle f, \Phi\rangle$ is valid at the same time because it is not allowed me to use Helly's theorem again. – Travis Wang Oct 18 '14 at 15:35
  • Many many thanks to you. I have known all the details of your proof now. If you have been bothered by me, I am deeply sorry. – Travis Wang Oct 20 '14 at 13:53
  • That's fine, I am glad you understand the proof. I just couldn't see where exactly you had been struggling. Sorry about that. – Tomasz Kania Oct 20 '14 at 14:00